NIMCET Vector Previous Year Questions (PYQs) – Page 1 of 10
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Solution
Solution
Given:
\( \vec{a} = \hat{i} - \hat{k}, \quad \) \(\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k},\) \( \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \)
Form the matrix:
\( M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix} \)
Find the determinant:
\( \det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1 \)
Since the determinant is constant and non-zero, the vectors are linearly independent.
\( \boxed{\text{The matrix does not depend on } x \text{ or } y} \)
Solution
Solution
Quick Solution
Given: \( \vec{a}, \vec{b} \) are unit vectors and
\( 2\vec{a} + \vec{b} = 3 \)
Take magnitude on both sides:
\( |2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9 \)
Use identity:
\[ |2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]
Set equal to 9:
\[ 5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ \]
Solution
Solution
Vector Projection Problem
Given: A vector of magnitude 5 makes equal angles with x, y, and z axes.
To Find: Sum of magnitudes of projections on each axis.
Let angle with each axis be \( \alpha \). Then, from direction cosine identity: \[ \cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \Rightarrow 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}} \]
Projection on each axis: \( 5 \cdot \frac{1}{\sqrt{3}} \)
Sum = \( 3 \cdot \frac{5}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \boxed{5\sqrt{3}} \)
✅ Final Answer: \( \boxed{5\sqrt{3}} \)
Solution
Expand
\[|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 2 - 2\vec{a}\cdot\vec{b}\]
\[|\vec{b}-\vec{c}|^2 = 2 - 2\vec{b}\cdot\vec{c}\]
\[|\vec{c}-\vec{a}|^2 = 2 - 2\vec{c}\cdot\vec{a}\]
Add:
\[= 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\]
Find Maximum :
\[|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0\]
\[3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0\]
\[\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \geq -\frac{3}{2}\]
Final Answer:
\[6 - 2 \times \left(-\frac{3}{2}\right) = 6 + 3 = \boxed{9}\]
✅ Expression does not exceed 9
Solution
Solution
Solution
Given: $\alpha (2\vec a-\vec b)+\beta (\vec a+2\vec b)=8\vec b-\vec a$ for all vectors $\vec a,\vec b$.
Combine like terms: $(2\alpha+\beta)\vec a+(-\alpha+2\beta)\vec b=-\vec a+8\vec b$.
Equate coefficients:
$2\alpha+\beta=-1$,
$-\alpha+2\beta=8$.
Solve:
From $2\alpha+\beta=-1 \Rightarrow \beta=-1-2\alpha$.
Substitute in $-\alpha+2\beta=8$:
$-\alpha+2(-1-2\alpha)=8 \Rightarrow -\alpha-2-4\alpha=8 \Rightarrow -5\alpha=10 \Rightarrow \alpha=-2$.
Then $\beta=-1-2(-2)=3$.
Answer: $\alpha=-2,\ \beta=3$.
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